Exercise 7.2 — Question 4

This question applies the Binomial Theorem to practical problems including finding middle terms, approximate values, remainders, and last digits of large powers.

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Key Formula Recap

The general term of the binomial expansion $(a+b{)}^n$ is:

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r,r=0,1,2,\dots ,n$

The total number of terms in the expansion is $n+1$.

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Middle Terms

To find the middle term(s) of $(a+b{)}^n$:

• If $n$ is even: there is one middle term, $T_{\frac{n}{2}+1}$.
• If $n$ is odd: there are two middle terms, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$.

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Approximate Values Using the Binomial Theorem

Expressions like $(1+x{)}^n$ where $x$ is small can be approximated by taking the first few terms of the expansion.

Example: Find an approximate value of $(1.02{)}^{10}$.

Write $1.02=1+0.02$, so:

$(1.02{)}^{10}=(1+0.02{)}^{10}={\sum }_{r=0}^{10}\left(\genfrac{}{}{0ex}{}{10}{r}\right)(0.02{)}^r$

Taking the first four terms:

$\approx 1+10(0.02)+45(0.0004)+120(0.000008)$ $=1+0.2+0.018+0.00096\approx 1.219$

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Finding Remainders Using the Binomial Theorem

To find the remainder when a large power is divided by a number, rewrite the base as $(\text{multiple of divisor}\pm 1{)}^n$.

Example: Find the remainder when $7^{100}$ is divided by $6$.

Write $7=6+1$, so:

$7^{100}=(6+1{)}^{100}={\sum }_{r=0}^{100}(\genfrac{}{}{0ex}{}{100}{r})6^r\cdot 1^{100-r}$

Every term with $r\ge 1$ contains a factor of $6$, so:

$7^{100}=\left(\genfrac{}{}{0ex}{}{100}{0}\right)+6k=1+6k\text{for some integer }k$

Therefore, the remainder is 1 when $7^{100}$ is divided by $6$.

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Finding the Last Digit Using the Binomial Theorem

The last digit (units digit) of a large power is the remainder when divided by $10$.

Example: Find the last digit of $3^{100}$.

Note that $3^4=81$, so $3^{100}=(3^4{)}^{25}=81^{25}$.

Write $81=80+1$:

$81^{25}=(80+1{)}^{25}={\sum }_{r=0}^{25}(\genfrac{}{}{0ex}{}{25}{r})80^r$

Every term with $r\ge 1$ contains a factor of $80$ (hence a factor of $10$), so:

$81^{25}\equiv 1(\mathrm{mod}10)$

The last digit of $3^{100}$ is 1.